为什么不两者都做？
在每个端点处画一个圆圈，并在两者之间画一条线。
编辑rof，只是无法阻止自己。
实际上，您不想使用pygame.draw.line，因为它会作弊。它填充 1 像素宽的行或列（取决于攻角）像素。如果您确实以大致垂直的角度，0 度或 90 度，这不是问题，但在 45 度时，您会注意到一种串豆效应。
唯一的解决方案是在每个像素的距离处画一个圆圈。这里...
import pygame, random
screen = pygame.display.set_mode((800,600))
draw_on = False
last_pos = (0, 0)
color = (255, 128, 0)
radius = 10
def roundline(srf, color, start, end, radius=1):
dx = end[0]-start[0]
dy = end[1]-start[1]
distance = max(abs(dx), abs(dy))
for i in range(distance):
x = int( start[0]+float(i)/distance*dx)
y = int( start[1]+float(i)/distance*dy)
pygame.draw.circle(srf, color, (x, y), radius)
try:
while True:
e = pygame.event.wait()
if e.type == pygame.QUIT:
raise StopIteration
if e.type == pygame.MOUSEBUTTONDOWN:
color = (random.randrange(256), random.randrange(256), random.randrange(256))
pygame.draw.circle(screen, color, e.pos, radius)
draw_on = True
if e.type == pygame.MOUSEBUTTONUP:
draw_on = False
if e.type == pygame.MOUSEMOTION:
if draw_on:
pygame.draw.circle(screen, color, e.pos, radius)
roundline(screen, color, e.pos, last_pos, radius)
last_pos = e.pos
pygame.display.flip()
except StopIteration:
pass
pygame.quit()
在每个循环步骤中不进行 blitting 可以提高绘图速度（使用从前一个改编的代码可以消除我机器上的滞后问题）
import pygame, random
screen = pygame.display.set_mode((800,600))
draw_on = False
last_pos = (0, 0)
color = (255, 128, 0)
radius = 10
def roundline(srf, color, start, end, radius=1):
dx = end[0]-start[0]
dy = end[1]-start[1]
distance = max(abs(dx), abs(dy))
for i in range(distance):
x = int( start[0]+float(i)/distance*dx)
y = int( start[1]+float(i)/distance*dy)
pygame.display.update(pygame.draw.circle(srf, color, (x, y), radius))
try:
while True:
e = pygame.event.wait()
if e.type == pygame.QUIT:
raise StopIteration
if e.type == pygame.MOUSEBUTTONDOWN:
color = (random.randrange(256), random.randrange(256), random.randrange(256))
pygame.draw.circle(screen, color, e.pos, radius)
draw_on = True
if e.type == pygame.MOUSEBUTTONUP:
draw_on = False
if e.type == pygame.MOUSEMOTION:
if draw_on:
pygame.display.update(pygame.draw.circle(screen, color, e.pos, radius))
roundline(screen, color, e.pos, last_pos, radius)
last_pos = e.pos
#pygame.display.flip()
except StopIteration:
pass
pygame.quit()
对于第一个问题，您需要有一个背景，即使它只是一种颜色。我制作的复制乒乓球游戏也遇到了同样的问题。下面是我做的一个复制画图程序的例子，左键绘制，右键擦除，点击彩色图像选择颜色，向上键清屏：
import os
os.environ['SDL_VIDEO_CENTERED'] = '1'
from pygamehelper import *
from pygame import *
from pygame.locals import *
from vec2d import *
from math import e, pi, cos, sin, sqrt
from random import uniform
class Starter(PygameHelper):
def __init__(self):
self.w, self.h = 800, 600
PygameHelper.__init__(self, size=(self.w, self.h), fill=((255,255,255)))
self.img= pygame.image.load("colors.png")
self.screen.blit(self.img, (0,0))
self.drawcolor= (0,0,0)
self.x= 0
def update(self):
pass
def keyUp(self, key):
if key==K_UP:
self.screen.fill((255,255,255))
self.screen.blit(self.img, (0,0))
def mouseUp(self, button, pos):
pass
def mouseMotion(self, buttons, pos, rel):
if pos[1]>=172:
if buttons[0]==1:
#pygame.draw.circle(self.screen, (0,0,0), pos, 5)
pygame.draw.line(self.screen, self.drawcolor, pos, (pos[0]-rel[0], pos[1]-rel[1]),5)
if buttons[2]==1:
pygame.draw.circle(self.screen, (255,255,255), pos, 30)
if buttons[1]==1:
#RAINBOW MODE
color= self.screen.get_at((self.x, 0))
pygame.draw.line(self.screen, color, pos, (pos[0]-rel[0], pos[1]-rel[1]), 5)
self.x+= 1
if self.x>172: self.x=0
else:
if pos[0]<172:
if buttons[0]==1:
self.drawcolor= self.screen.get_at(pos)
pygame.draw.circle(self.screen, self.drawcolor, (250, 100), 30)
def draw(self):
pass
#self.screen.fill((255,255,255))
#pygame.draw.circle(self.screen, (0,0,0), (50,100), 20)
s = Starter()
s.mainLoop(40)
这是 Matthew示例的简化版本，遗憾的是它无法运行。
当鼠标移动时，pygame.MOUSEMOTION事件被添加到包含位置和相对移动的事件队列中。您可以使用这些来计算先前的位置，然后将两个点传递给pygame.draw.line。
pygame.MOUSEMOTION事件还有一个buttons 属性，您可以使用它来检查当前按下的鼠标按钮。
import os
import random
import pygame as pg
class App:
def __init__(self):
os.environ['SDL_VIDEO_CENTERED'] = '1'
pg.init()
self.w, self.h = 800, 600
self.screen = pg.display.set_mode((self.w, self.h))
self.screen.fill(pg.Color('white'))
self.clock = pg.time.Clock()
self.drawcolor = (0, 0, 0)
def mainloop(self):
while True:
for event in pg.event.get():
if event.type == pg.QUIT:
return
elif event.type == pg.MOUSEBUTTONDOWN:
if event.button == 2: # Color picker (middle mouse button).
self.drawcolor = self.screen.get_at(pos)
# Pick a random color.
# self.drawcolor = [random.randrange(256) for _ in range(3)]
elif event.type == pg.MOUSEMOTION:
pos, rel = event.pos, event.rel
if event.buttons[0]: # If the left mouse button is down.
# Draw a line from the pos to the previous pos.
pg.draw.line(self.screen, self.drawcolor, pos, (pos[0]-rel[0], pos[1]-rel[1]), 5)
elif event.buttons[2]: # If the right mouse button is down.
# Erase by drawing a circle.
pg.draw.circle(self.screen, (255, 255, 255), pos, 30)
pg.display.flip()
self.clock.tick(30)
if __name__ == '__main__':
app = App()
app.mainloop()
pg.quit()
我制作了一个更有效的圆形功能版本。
import pygame as pg,random,math
screen = pg.display.set_mode((800,600))
draw_on = False
last_pos = (0, 0)
color = (255, 128, 0)
radius = 10
def roundline(srf, color, start, end, radius=1):
pg.display.update([drawline(srf,color,start,end,radius),
pg.draw.circle(srf, color, end, radius),
pg.draw.circle(srf, color, start, radius)])
def drawline(srf, color,start,end,width):
v=(start[0]- end[0],start[1]- end[1])
betrag=width/math.sqrt(v[0]*v[0]+v[1]*v[1])
v_=(v[1]*betrag,-betrag*v[0])
return pg.draw.polygon(srf, color,(
(int(start[0]+v_[0]),int(start[1]+v_[1])),
(int(start[0]-v_[0]),int(start[1]-v_[1])),
(int(end[0]-v_[0]),int(end[1]-v_[1])),
(int(end[0]+v_[0]),int(end[1]+v_[1]))
))
try:
while True:
e = pg.event.wait()
if e.type == pg.QUIT:
raise StopIteration
elif e.type == pg.MOUSEBUTTONDOWN:
color = (random.randrange(256), random.randrange(256), random.randrange(256))
#pg.draw.circle(screen, color, e.pos, radius)
draw_on = True
start=e.pos
elif e.type == pg.MOUSEBUTTONUP:
draw_on = False
elif e.type == pg.MOUSEMOTION:
if draw_on:
roundline(screen, color, e.pos, last_pos, radius)
last_pos = e.pos
elif e.type == pg.MOUSEWHEEL:
radius=max(1,radius+e.y)
finally:
pg.quit()